Planetary motion considering natural satellite

Using Kepler's laws of gravitation we can determine the time period of a planet revolving around the sun. However, this excludes the gravitational effect due to satelite(s) orbiting around the planet.

How much difference would this create in the time period of a planet?

For a simple model of a planet of mass $$M$$, Radius $$R$$ having a natural satellite of mass $$m$$ at a distance $$ro$$ from the centre of the planet, revolving around it. The planet revolves around a sun of mass $$M_s$$ at a distance of $$r$$ from the sun. Assuming all orbits to be circular and the orbital planes of earth around sun, and planet around the earth to be perpendicular, is it possible to calculate the time period of revolution of the planet around the sun?

• I believe the Earth-Moon system has the least difference in mass between the planet and its satellite, and the moon is only about 1% of the mass of the Earth. Assuming circular orbits is probably a bigger error than neglecting the mass of the moon when calculating the Earth's orbital period. May 23, 2020 at 16:55
• Oh I see thanks. Would this always be true even in the hypothetical case in the Question, if $M_s >> M >> m$\$? May 23, 2020 at 16:58

If you use the center of mass of the planet+moons, as well as their total mass, there is no issue. That CM will revolve according to Kepler's laws. (As a virtual fellow student of his, I must insist on correct spelling!)

How far ahead/behind the planet will actually be with respect to that CM will depend on the details and can vary from one revolution to the next, depending on where the moon ends up.

• Wow that's really interesting! I thought it would be a combination of three "double star" systems, and obtained three CMs and I wasn't really sure what would happen! I guess considering sun to be stationary is the best way out May 23, 2020 at 17:09
• Well, it's still true that the sun is not stationary, only the CM of all three is. It's just easier to think of planet+moon as one combined object first.
– user257090
May 23, 2020 at 17:31
• @user600016 You can see the motion of the solar system's centre of mass relative to the Sun here: physics.stackexchange.com/a/546072/123208 May 23, 2020 at 19:22

One star, one planet, and one moon is the Three-Body Problem, which used to be one of the things famous physicists like Lagrange and Poincaré studied for significant fractions of their careers. The problem has no closed-form solution, but the search for one inspired the invention of significant pieces of mathematics in analysis, topology, chaos theory, and more. It is still studied today, using the tools developed along the way (especially numerical integration of systems of partial differential equations), but it no longer plays the central role in physics that it did two centuries ago. Many different special cases have been solved, depending on what is assumed about ratios of masses, ratios of distances, angular momentum, and whether the motion is confined to a plane.

In the "restricted" three-body problem, we assume the mass of the third object is negligible, in the sense that the star and planet orbit each other in a Keplerian two-body orbit, and their gravity controls the moon's orbit, but the moon's mass does not affect the orbit of the star and planet about each other. This kind of relation actually holds much better for artificial satellites than natural ones. For example, the ratio of our sun's mass to its major satellites ranges from 10^3 for Jupiter to 10^7 for Mercury, or 10^8 for Pluto. The ratio of planets to their major moons is mostly 10^5; Earth's Moon is relatively very large, with a mass ratio of only 80. The ratio of the Moon's mass to an artificial satellite's is 10^20 for a 700 kg vehicle.

Specializing to circular orbits in a fixed plane, but not necessarily negligible third mass, we find the five Lagrange points, three of which are due to Euler. The two Lagrange himself found, L4 and L5, are the locations where the third mass forms an equilateral triangle with the other two masses. Treating each planet plus its moons as a single mass, the Sun and planet can form three-body systems with things at their L4 and L5, and those orbits would be stable if there weren't many more than three significant bodies. Even with all the other planets, moons, and smaller bodies in play, there are in fact long-term stable communities of "trojan" asteroids that are found in the L4 and L5 of the Sun-Jupiter system, the Sun-Neptune system, and others; there are even L4/L5 trojan relations among the moons of the Saturn system.

Please note, these are only "points" when viewed in a special coordinate system rotating about the center of mass with a particular angular velocity. In any other frame, the orbits of these "points" become much more complicated. That particular angular velocity, however, tells us something very interesting. That is, if we consider each of the second and the third masses to be in separate two-body orbits in the usual way (that is, assuming only the Sun exerts gravitational force on all of the other bodies), Kepler's Third Law would tell us that whichever body was farthest away from the center of mass (the Sun, because we already assumed everything else was effectively zero) would take the most time to complete its orbit. For bodies at the five Lagrange points, however, this is not true! All three objects turn at the same angular velocity, namely that of the rotating frame in which they can be viewed as points. This is true for L4 and L5 even if all three masses are equal, or any other ratio. In that case, the three masses are at the corners of an equilateral triangle, and the whole triangle spins around its center of mass as if it were a single, solid object. The analogous equation to Kepler's is $$\omega^2=G(m_1+m_2+m_3)/D^3$$, where D is the distance from the three masses to each other. The largest mass is closest to the center of mass, so it makes the smallest circle, and the smallest mass is farthest from the center of mass, so it makes the largest circle, but all three masses revolve in their orbit at the same angular speed.